#1




Which would you prefer: A half a barrel of dimes or a barrel of nickels?
Is there a way to mathmatically determine the answer to the question: Which would you prefer: A half a barrel of dimes or a barrel of nickels?
It seems to me that there isn't enough information. But I haven't had any mathmatics in many years. 
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#2




Stacking geometry notwithstanding, you can answer this problem on the basis of the relative volume of the two coins. Assuming that the barrel referred to in each case is of the same volume, look a a dime, and then a nickel  you will notice that the volume of a dime is considerably less than half of the volume of a nickel. When you fill a barrel half full of dimes, it stands to reason that the value of these coins will be greater than a full barrel full of nickels.

#3




Dimes and quarters, on the other hand...

#4




I'll take a halfbarrel that is halffull halfway right on up to the halfway point with dimes, but only if half of it's filled.

#5




Dimes, I think.

#6




Dimes, most obviously  Not only can you get at least twice as many in the same volume, they're twice as valuable coinforcoin, meaning that the barrel of dimes would be at least four times more valuable than the nickels.

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The dime measured 0.796" in diameter and 0.050" thick. The nickel measured 0.838" in diameter and 0.075" thick. Since cylinders don't pack well, I calculated volume as if the surface was a square rather than a circle. (This is where I may be off, is the packing better if they're arranged in "hexagons"). The dime's "volume" is 0.0316 in^3 and the nickel's "volume" is 0.0527 in^3. Since the dime's volume is greater than half the nickel's volume, but it is only twice the value, I think the better choice is the full barrel of nickels. 
#8




Interesting that everyone else has chosen dimes while I've been off measuring...
It seems that relative volumes on this scale are difficult to estimate. 
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1. A Full barrel of nickels or 2. A HALF barrel of dimes. I'd still go with the dimes. 
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Of course, that's only if you're interested in maximizing cash value. If your personal preferences depend on other criteria, that's your decision. 
#12




This reminds me of a joke I just read:
There's a little fellow named Junior who hangs out at the local grocery store. The manager doesn't know what Junior's problem is, but the boys like to tease him. They say he is two bricks short of a load, or his elevator doesn't go all the way to the top. To prove it, sometimes the boys offer Junior his choice between a nickel and a dime. He always takes the nickel, they say, because it's bigger. One day after Junior grabbed the nickel, the store manager got him off to one side and said, "Junior, those boys are making fun of you. They think you don't know the dime is worth more than the nickel. Are you grabbing the nickel because it's bigger, or what?" Junior said, "Well, if I took the dime, they'd quit doing it!" 
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At worst, you'd have half as many dimes as nickels (same monetary value), and more likely you'd have at least some extra dimes. And the halfbarrel of dimes would be lighter  easier to carry to the bank. Now, if you'll offer me a halfbarrel of paper money, I'll forget I ever heard about the dimes. 
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#16




JTC: Since cylinders don't pack well, I calculated volume as if the surface was a square rather than a circle. (This is where I may be off, is the packing better if they're arranged in "hexagons").
Yes, it is. The hexagonal lattice packing (with each circle surrounded by six circles touching it) is the maximally dense packing of uniformlysized circles. (Math Phun Phact: The corresponding statement for three dimensions, that hexagonal lattice packing of uniformlysized spheres is maximally dense, was conjectured by Johannes Kepler back in 1611 but not actually proved until a few years ago, by Thomas Hales. Physics Phun Phact: some dude whose name I don't remember experimentally tested the Kepler conjecture by leaving dried peas in a pot of water overnight and noting that their expansion had squeezed them into little dodecahedrons, suggesting that the spacesaving distribution they naturally settled into was the hexagonal lattice packing. This illustrates why the Kepler conjecture always used to be introduced with the words "Many mathematicians believe, and all physicists know, that..." I always get a chuckle out of thinking of that guy counting the little faces on a soggy pea. ) 
#17




Weigh the barrels. Let w_{1} be the weight of the barrel of nickels, and w_{2} be the weight of the barrel of dimes. Let d be the weight of a dime, and n the weigh of a nickel.
The value of the barrel of dimes is w_{1}/10d, and the value of the barrel of nickels is w_{2}/5n. If w_{1}/w_{2} > 2d/n, then the barrel of dimes is worth more. Otherwise, it's the barrel of nickels. 
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#19




It's not mathematics, it's just simple logic.
Assume that dimes and nickels are exactly the same shape. Fill half a barrel with dimes, and fill a barrel with nickels. By definition, the two are worth the exact same value the barrel of nickels has twice as many coins, but each coin is worth half the value of a dime. But, because nickels are larger than dimes, there must be *more* dimes in the halfbarrel than half the number of nickels in the full barrel. Therefore, the halfbarrel of dimes is worth more. 
#20




chrisk: I don't believe that dodecahedrons will pack together in a hexagonal lattice... or any lattice very neatly. I could be wrong.
I don't think you're wrong about regular dodecahedra, but the ones that hexagonallatticepacked spheres expand into are rhombic dodecahedra, which are spacefilling: Quote:

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I'm not quite sure how I managed to be off by .090" Might as well be a mile. Quote:
So I guess I should have taken the half barel after all. I guess that's the difference between precision and accuracy... 
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Given that dimes and nickels are the same overall shape (highly flattened cylinders,) I think you can make a convincing argument that there are at least as many dimes as half the nickels, and therefore that the dimes are at least as valuable. To prove anything further, you kinda need to go to the maths. Quote:

#23




Actually, to express it mathematically:
Let VB= The volume of the Barrel. Let VD= The volume of a dime. Let VN= The volume of a nickel. The question is the relationship between 1/2*(VB/VD)*10 and (VB/VN)*5. which can be reduced to (VB/VD)*5 vs. (VB/VN)*5 which goes down to (VB/VD) vs. (VB/VN) which can further be reduced to (1/VD) vs. (1/VN). Since VD<VN, 1/VD>1/VN. 
#24




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I'd still take dimes  I'm betting they're less than half the volume of a nickel, and even if they aren't, they're still easier to move about. 
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15 gallons of dimes: 157,500 dimes = $15,750. Take the dimes. 
#26




I'm with John on this. Unless I'm totally missing something. Which is certainly possible. I don't know how to write a math formula for it, but logic says the dimnes win.
Pretend that nickles and dimes are exactly the same size coin. Pretend that a full barrel hold 1,000 coins (that's one small barrel, but bear with me). A full barrel of nickels would be 1,000 coins at $0.05 each, or $50. A half barrel of dimes would be 500 coins at $0.10 each, or $50. So only in the case of the two coins being the same size would the value of both barrels be the same. But dimes are clearly much smaller than nickles, so a half barrel of dimes would be much more than 500 coins. Which means that the half barrel would have a much higher value. (If I missed something, go easy on me. I ain't had no fancy book larnin') 
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